parabola y=ax^2+k has vertex (0,-1) and passes through the point (3,3). find the equation

$\text{if we've got vertex at }(0,-1) \text{ then}\\ y = a x^2-1\\ \text{we're told that }(3,3) \text{ is on the parabola}\\ 3 = a 3^2 - 1\\ 4 = 9a\\ a = \dfrac{4}{9}\\ y = \dfrac 4 9 x^2 - 1$